Problem: Solve for $x$ and $y$ using elimination. ${2x+5y = 27}$ ${-2x+4y = 0}$
Explanation: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Add the equations together. Notice that the terms $2x$ and $-2x$ cancel out. $9y = 27$ $\dfrac{9y}{{9}} = \dfrac{27}{{9}}$ ${y = 3}$ Now that you know ${y = 3}$ , plug it back into $\thinspace {2x+5y = 27}\thinspace$ to find $x$ ${2x + 5}{(3)}{= 27}$ $2x+15 = 27$ $2x+15{-15} = 27{-15}$ $2x = 12$ $\dfrac{2x}{{2}} = \dfrac{12}{{2}}$ ${x = 6}$ You can also plug ${y = 3}$ into $\thinspace {-2x+4y = 0}\thinspace$ and get the same answer for $x$ : ${-2x + 4}{(3)}{= 0}$ ${x = 6}$